Thread: Bounds on integration. Have answer, need clarification.

Evaluate the iterated integral by converting to polar coordinates.




I know how to do the integration part; that's not my question. My question is how the original solver of the problem came up with the bounds of integration: the first integral from 0 to pi/4 and the second from 0 to sqrt(2).

I came up with the right answer, but I think I get it using the wrong methods. I took x=sqrt(2-y^2), turned that into x^2+y^2 = 2, giving me a radius of sqrt(2), and therefore 0 <= r <= sqrt(2). For the theta however, I'm a little lost. Is it obtained simply by observing the fact that x=y makes a 45 degree angle with both axes?

Any help is greatly appreciated.

Where do and intersect?

Just follow the limits of integration. The first integral is y: from 0 to 1. Draw it. The second integral is x: from y to sqrt(2-y^2) as y varies from 0 to 1. Again, draw it.

You will get a quarter of a circle, in the first quadrant, bisected along a 45 degree angle by the curve y = x. The region of integration is the bounded region below the bisector.

You can describe this region in polar coordinates quite easily. The quarter-circle is described by the polar function r = sqrt(2). The curve y = x can also be represented as a polar function theta = pi/4 (45 degrees). If you did not know the function beforehand, you could have simply transformed each rectangular equation into a polar equation, then simplified/solved until the resulting polar equation is recognizable; for example: x = y => rcos(t) = rsin(t) => cos(t) = sin(t) => tan(t) = cot(t) = 1 => t = pi/4 = 45 degrees (where t = theta). I assume you can do the same for the circle.

Now that you have obtained corresponding descriptions in polar coordinates, you must setup the appropriate limits of integration. As theta now varies from 0 to pi/4, r must vary from 0 to sqrt(2) to "obtain" "cover" "get" whatever - the entire region.

Get use to doing such things; it will (perhaps) become worse in spherical and cylindrical coordinates.